Answer:
Final Velocity (Vf)= 139.864 ft/s
Time (t)= 4,34 s
Explanation:
This is a free fall problem, to solve it we will apply free fall concepts:
In a free fall the acceletarion is gravity (g) = 9,81 m/s2, if we convert it to ft/s^2 = g= 32.174 ft/s^2
Final velocity is Vf= Vo+ g*t[tex]Vf^{2} = Vo^{2} +2*g*hwhere h is height (304 ft in this case).
Vo =0 since the hammer wasn't moving when it stared to fall
Then Vf^2= 0 + 2* 32.174 ft/s^2 *304 ft
Vf^2= 19,561.8224 ft^2/s^2
Vf=[sqrt{19561.8224 ft^2/s^2}
Vf=139.864 ft/s
Time t= (Vf-Vo)/g => (139.864 ft/s-0)/32.174 ft/s^2 = 4.34 sec
Good luck!
The Torricelli's theorem states that the (velocity—pressure-density) of liquid flowing out of an orifice is proportional to the square root of the (height-pressure-velocity) of liquid above the center of the orifice.
Answer:
The correct answer is 'velocity'of liquid flowing out of an orifice is proportional to the square root of the 'height' of liquid above the center of the orifice.
Explanation:
Torricelli's theorem states that
[tex]v_{exit}=\sqrt{2gh}[/tex]
where
[tex]v_{exit}[/tex] is the velocity with which the fluid leaves orifice
[tex]h[/tex] is the head under which the flow occurs.
Thus we can compare the given options to arrive at the correct answer
Velocity is proportional to square root of head under which the flow occurs.
Explain the difference between planning and shaping by the help of sketch
Explanation:
In shaping work piece will be stationary and tool will reciprocates,but on the other hand in planning work piece will reciprocates and tool will be stationary.Shaping is used for small work piece and planning is used for large work piece.Both shaping and planning are not continuous cutting process,cutting action take place in forward stroke and return stroke is idle stroke.The velocity of return stroke is much more than the forward stroke.
Derive the dimensions of specific heat that is defined as the amount of heat required to elevate the temperature of an object of mass 1 kg by 1 degree Celcius.
Answer:
Dimension of specific heat will be [tex]=L^2T^{-2}\Theta ^{-1}[/tex]
Explanation:
We know that heat [tex]Q=mc\Delta T[/tex], Q is heat generated, m is mass, c is specific heat and [tex]\Delta T[/tex] is temperature difference
From formula we can write [tex]c=\frac{Q}{m\times \Delta T}[/tex]
Now unit of Q is joule or N-m
Newton can be written as [tex]kgm/sec^2[/tex]
So unit of Q is [tex]kgm^2/sec^2[/tex]
For dimension we use M for kg, L for meter(m) ,T for sec and [tex]\Theta[/tex] for temperature
So dimension of Q is [tex]ML^2T^{-2}[/tex]
So dimension of specific heat will be [tex]\frac{ML^2T^{-2}}{M\Theta }=L^2T^{-2}\Theta ^{-1}[/tex]
The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The absolute pressure at a depth of 5 m in liquid whose specific gravity is 0.8 at the same location.
Answer:
a)Patm=135.95Kpa
b)Pabs=175.91Kpa
Explanation:
the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation
Pabs=Pw+Patm(1)
Where
Pabs=absolute pressure
Pw=Water pressure
Patm= atmospheric pressure
Water pressure is calculated with the following equation
Pw=γ.h(2)
where
γ=especific weight of water=9.81KN/M^3
H=depht
A)
Solving using ecuations 1 y 2
Patm=Pabs-Pw
Patm=185-9.81*5=135.95Kpa
B)
Solving using ecuations 1 y 2, and atmospheric pressure
Pabs=0.8x5x9.81+135.95=175.91Kpa
A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates with a maximum amplitude of 10 mm. What is its natural frequency?
Answer:
f=1.59 Hz
Explanation:
Given that
Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.
Velocity = 100 mm/s
Maximum amplitude = 10 mm
We know that for a simple undamped system spring mass system
[tex]V_{max}=\omega A[/tex]
now by putting the values
[tex]V_{max}=\omega A[/tex]
100 = ω x 10
ω = 10 rad/s
We also know that
ω=2π f
10 = 2 x π x f
f=1.59 Hz
So the natural frequency will be f=1.59 Hz.
A 4-kg-plastic tank that has a volume of 0.2 m^3 is filled with liquid water. Assuming the density of water is 1000 kg/m^3, determine the weight the combined system.
Answer:
The weight of the combined system is 2001.24 Newtons
Explanation:
From the basic relation between mass, density and volume we know that
[tex]density=\frac{Mass}{Volume}[/tex]
In our context we are given that the density of water is 1000 kg per cubic meters
Thus we can find the mass of 0.2 cubic meters of water using the above relation as
[tex]1000kg/m^{3}=\frac{Mass}{0.2m^{3}}\\\\\therefore Mass=1000kg/m^{3}\times 0.2m^{3}=200kg[/tex]
Hence the mass of water in the tank is 200 kilograms.
The total mass of water and the plastic tank thus becomes
[tex]200+4=204kg[/tex]
Now we know that weight of any given mass is calculated as
[tex]Weight=mass\g[/tex]
where,
'g' is the acceleration due to gravity with value = [tex]9.81m/s^{2}[/tex]
Applying the values in the above equation we get
[tex]Weight=204\times 9.81=2001.24Newtons[/tex]
The pulley has mass 12.0 kg, outer radius Ro=250 mm, inner radius Ri=200 mm, and radius of gyration kO=231 mm. Cylinder A weighs 71.0 N. Assume there is no friction between the pulley and its axle and that the rope is massless. At the instant when ω=69.0 rad/s clockwise, what is the kinetic energy of the system?
The total kinetic energy of the system is approximately 17128.26 J.T
What is the energy?
Calculate the moment of inertia (I) of the pulley:
= 0.5 * mass * (outer radius² + inner radius²)
= [tex]0.5 * 12.0 kg * ((0.250 m)² + (0.200 m)²)[/tex]
= 1.925 kg * m²
Use the parallel-axis theorem to find I:
I = [tex]1.925 kg * m² + 12.0 kg * (0.231 m)²[/tex]
I = 2.5683 kg * m²
Calculate the kinetic energy of the pulley:
= 0.5 * I * omega²
= [tex]0.5 * 2.5683 kg * m² * (69.0 rad/s)[/tex]
= 6555.63 J
Calculate the linear velocity of cylinder A:
v = outer radius * omega
v =[tex]0.250 m * 69.0 rad/s[/tex]
v = 17.25 m/s
Calculate the kinetic energy of cylinder A:
= 0.5 * mass * v²
= [tex]0.5 * 71.0 kg * (17.25 m/s)²[/tex]
= 10572.63 J
KEtotal = KEpulley + KEcylinder
= [tex]6555.63 J + 10572.63 J[/tex]
= 17128.26 J
What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ*dV/dy? Then, by dimensional analysis, show that the shear stress has the same units as momentum divided by (area*time).What are the unit or dimensions of viscosity?
Answer:
1) Dimensions of shear rate is [tex][T^{-1}][/tex] .
2)Dimensions of shear stress are [tex][ML^{-1}T^{-2}][/tex]
Explanation:
Since the dimensions of velocity 'v' are [tex][LT^{-1}][/tex] and the dimensions of distance 'y' are [tex][L][/tex] , thus the dimensions of [tex]\frac{dv}{dy}[/tex] become
[tex]\frac{[LT^{-1}]}{[L]}=[T^{-1}][/tex] and hence the units become [tex]s^{-1}[/tex].
Now we know that the dimensions of coefficient of dynamic viscosity [tex]\mu [/tex] are [tex][ML^{-1}T^{-1}][/tex] thus the dimensions of shear stress can be obtained from the given formula as
[tex][\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}][/tex]
Now we know that dimensions of momentum are [tex][MLT^{-1}][/tex]
The dimensions of [tex]Area\times time[/tex] are [tex][L^{2}T][/tex]
Thus the dimensions of [tex]\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}][/tex]
Which is same as that of shear stress. Hence proved.
1.19. A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m·s−2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?\
Answer:
a) 19094 N
b) 110.055 kPa
c) 1222 J
Explanation:
The force on the gas is the weight plus the atmospheric pressure multiplied by the piston area
F = P + p * A
F = m * g + p * π/4 * d^2
F = 150 * 9.813 + 101570 * π/4 * 0.47^2 = 19094 N
The pressure is the force divided by the area of the piston
p1 = F / A
p1 = F / (π/4 * d^2)
p1 = 19094 / (π/4 * 0.47^2) = 110055 Pa = 110.055 kPa
variation of gravitational potential energy is defined as
ΔEp = m * g * Δh
ΔEp = 150 * 9.813 * 0.83 = 1222 J
In this exercise we have to use the knowledge of force to calculate the required energies, so we have to:
a) 19094 N
b) 110.055 kPa
c) 1222 J
What is the concept of force?In the field of physics, force is a physical action that causes deformation or that changes the state of rest or movement of a given object.
a) Knowing that the force formula is defined by:
[tex]F = P + p * A\\F = m * g + p *\pi /4 * d^2\\F = 150 * 9.813 + 101570 * \pi /4 * 0.47^2 = 19094 N[/tex]
b) Knowing that the force exerted by an area is equal to the pressure in that area, we have:
[tex]p_1 = F / A\\p_1 = F / (\pi /4 * d^2)\\p_1 = 19094 / (\pi /4 * 0.47^2) = 110055 Pa = 110.055 kPa[/tex]
c)So calculating the potential energy we have:
[tex]\Delta E_p = m * g * \Delta h\\\Delta E_p = 150 * 9.813 * 0.83 = 1222 J[/tex]
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Find the error in the following proof that 2 = 1. Consider the equation a = b. Multiply both sides by a to obtain a 2 = ab. Subtract b 2 from both sides to get a 2 − b 2 = ab − b 2 . Now factor each side, (a + b)(a − b) = b(a − b), and divide each side by (a − b) to get a + b = b. Finally, let a and b equal 1, which shows that 2 = 1.
Answer:
You can't divide by zero
Explanation:
The error appears when you divide each side by (a - b). If a = b, then (a - b) = 0 and you can't divide each side by 0. Moreover, the equation before division, that is, (a + b)(a − b) = b(a − b) is true after replacing (a - b) = 0 because it gives 0 = 0.
The error in the proof lies in the step where we divided both sides by (a - b) . Since [tex]\( a = b \), \( (a - b) \)[/tex] becomes 0, making the division by 0 undefined.
The error occurs when dividing both sides by (a - b) . Since ( a = b ), ( (a - b) becomes 0. Division by 0 is undefined, leading to the invalid conclusion.
To elaborate, dividing both sides by (a - b) in the equation (a + b)(a - b) = b(a - b) results in:
[tex]\[ \frac{(a + b)(a - b)}{(a - b)} = \frac{b(a - b)}{(a - b)} \]\[ \frac{(a + b)\cancel{(a - b)}}{\cancel{(a - b)}} = \frac{b\cancel{(a - b)}}{\cancel{(a - b)}} \]\[ a + b = b \][/tex]
This step assumes (a - b) is not equal to zero, leading to the false conclusion that ( a + b = b ), which is only true when a and ( b ) are equal.
Therefore, the proof incorrectly concludes that ( 2 = 1 ) due to the division by zero, highlighting the importance of avoiding such mathematical errors.
A water tank is emptied through a pipe with an outlet 5m below the water surface level. What is the exit velocity? a) 2.1m/s b) 9.9 m/s c) -12.3m/s d)-4.8m/s e) 15.3m/s
Answer:
The correct answer is option 'b': 9.9 m/s
Explanation:
We know that for an ideal fluid the velocity of exit from a tank with the height of water 'h' is given by Torricelli's Law as
[tex]v=\sqrt{2gh}[/tex]
where,
'g' is acceleration due to gravity
'h' is the level of water
Applying the given values we obtain velocity as
[tex]v=\sqrt{2\times 9.81\times 5}=9.90m/s[/tex]
Convert 250 lb·ft to N.m. Express your answer using three significant figures.
Answer:
It will be equivalent to 338.95 N-m
Explanation:
We have to convert 250 lb-ft to N-m
We know that 1 lb = 4.45 N
So foe converting from lb to N we have to multiply with 4.45
So 250 lb = 250×4.45 =125 N
And we know that 1 feet = 0.3048 meter
Now we have to convert 250 lb-ft to N-m
So [tex]250lb-ft=250\times 4.45N\times 0.348M=338.95N-m[/tex]
So 250 lb-ft = 338.95 N-m
A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature of the steam at the turbine inlet is 400 oC, and the mass flow rate of steam through the cycle is 37 kg/s. Determine: a) the thermal efficiency of the cycle (%) and b) the net power output of the power plant (kW).
Answer:
a)31%
b)34MW
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
For this problem, we will first find the enthalpies in all states
h1=3231kJ/Kg
h2=2310kJ/Kg
h3=h4=272kJ/Kg
A) using the eficiency ecuation
Efficiency = (h1-h2) / (h1-h4)
Efficiency =(3231-2310)/(3231-272)=0.31=31%
b)using ecuation for Wout
Wout = m (h1-h2)
Wout=37(3231-2310)=34077KW=34.077MW
A substance temperature was 62 deg R. What is the temperature in deg C? A.) 50.7 B.) 45.54 C) 80.0 D) 94.4
Answer:
The temperature in degree Celsius will be -238.7055°C
Explanation:
We have given the substance temperature = 62°R
We have to convert degree Rankine to degree Celsius
For conversion from Rankine to Celsius we use formula
[tex]T_C=(T_R-491.67)\times\frac{5}{9}[/tex]
So [tex]T_C=(62-491.67)\times\frac{5}{9}[/tex]
[tex]T_C=-238.7055^{\circ}C[/tex]
So temperature in degree Celsius will be -238.7055°C
After calculation i got -238.7055°C but in option this is not given
The temperature of the substance will be "94.4°C". To understand the calculation, check below.
TemperatureAccording to the question,
Substance temperature, T°R = 62
or,
T°C = (T°R - 491.67) × [tex]\frac{5}{9}[/tex]
By substituting the values,
= -238.706
If we take the value,
T°C = (662 - 491.67) × [tex]\frac{5}{9}[/tex]
= 94.62°C or,
= 94.4°C
Thus the above response "Option D" is correct.
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An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [in J/m3 (which is equivalent to Pa)] given that it exhibits linear elastic stress-strain behavior.
Answer:
Modulus of resilience will be [tex]3216942.308j/m^3[/tex]
Explanation:
We have given yield strength [tex]\sigma _y=818MPa[/tex]
Elastic modulus E = 104 GPa
We have to find the modulus
Modulus of resilience is given by
Modulus of resilience [tex]=\frac{\sigma _y^2}{2E}[/tex], here [tex]\sigma _y[/tex] is yield strength and E is elastic modulus
Modulus of resilience [tex]=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3[/tex]
Give examples of engineering structures which can be modelled as thin walled cylinders.
Answer:
Pipes, pressure vessels, tanks, reactors, tubes and nozzles
Explanation:
Thin walled cylinders are typically defines as having wall thickness of 1/10 of the radius (doesn't matter much if inner or outer, they should be similar). Also this is used mostly for things that will be subject to some radial load, as opposed to axles and shafts.
As such, some structures that can be modeled as thin walled cylinders are pipes, pressure vessels, tanks, reactors, tubes and nozzles.
A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.
Answer:
1) Angle with x-axis = 42.03 degrees
2) Angle with y-axis =68.2 degrees
3) Angle with z-axis = 56.14 degrees
Explanation:
given any vector [tex]\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}[/tex]
and any x axis the angle between them is given by
[tex]\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}[/tex]
Angle between the vector and y axis is given by
[tex]\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}[/tex]
Similarly angle between z axis and the vector is given by
[tex]\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}[/tex]
The angles made by F will be "42.03°", "68.2°" and "56.14°".
Force and Vector:According to the question,
Force, F = 160 i + 80 j + 120 kN
Let any vector,
[tex]\vec r = x \hat i + y \hat j+ z \hat k[/tex]
The angle between x-axis be:
[tex]\Theta_x[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat i}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{x.i}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{160}{\sqrt{160^2+80^2+120^2} }[/tex])
= 42.03°
and,
The angle between y-axis be:
[tex]\Theta_y[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat j}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{y.i}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{80}{\sqrt{160^2+80^2+120^2} }[/tex])
= 68.2°
and,
The angle between z-axis be:
[tex]\Theta_z[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat k}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{z.k}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{120}{\sqrt{160^2+80^2+120^2} }[/tex])
= 56.145°
Thus the above approach is correct.
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The Air Force One (Boeing 747-200) has a long-range mission takeoff gross load 833,000 pounds (Ibm). What is Air Force One's takeoff mass in: a. gram (9) b. kilogram (kg) c. tonne (ton) d. Mton
Answer:
The mass in:
1) Grams =[tex]377765.5\times 10^{3}grams[/tex]
2) Kilograms =[tex]377765.5kilograms[/tex]
3)Tonnes =[tex]377.7655tonnes[/tex]
4) Megatonnes =[tex]0.378megatonnes[/tex]
Explanation: The given mass of the aircraft in pounds is 833,000 pounds.
Part 1)
Since we know that 1 pound equals 453.5 grams thus by ratio we have
833,000 pounds =[tex]833000lb\times 453.5\frac{g}{lb}=377765.5\times 10^{3}grams[/tex]
Part 2)
Since we know that 1000 grams equals 1 kilogram
thus the above mass in kilograms equals[tex]\frac{377765.5\times 10^{3}}{1000}=377765.5kg[/tex]
Part 3)
Since there are 1000 kilograms in 1 tonne
thus the given mass is converted into tonnes as
[tex]mass_{tonnes}=\frac{377765.5kg}{1000}=377.765tonnes[/tex]
Part 4)
Since 1 Mega tonne(Mton) equals 1000 tonnes thus the given mass is converted into mega tonnes as
[tex]mass_{M\cdot tonn}=\frac{377.7655tonnes}{1000}=0.378Megatonnes[/tex]
Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrust on the tyres.
Answer:
11.125°
Explanation:
Given:
Radius of bend, R = 100 m
Speed around the bend = 50 Km/hr = [tex]\frac{5}{18}\times50[/tex] = 13.89 m/s
Now,
We have the relation
[tex]\tan\theta=\frac{v^2}{gR}[/tex]
where,
θ = angle of banking
g is the acceleration due to gravity
on substituting the respective values, we get
[tex]\tan\theta=\frac{13.89^2}{9.81\times100}[/tex]
or
[tex]\tan\theta=0.1966[/tex]
or
θ = 11.125°
You live on a street that runs East to West. You just had 2 inche of snow and you live on the North side of the street. You return from class at 2PM and notice all the snow on your sidewalk is gone but across the street it is still there. No one removed the snow. How did it go away?
Answer:
The heat from the sun melted it
Explanation:
If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.
Yield strength (Sy) is typically defined as the point on the stress–strain curve with a strain of during a tension test. a) 0.1% b) 0.2% c) 0.3% c) 0.4%
Answer:
The correct answer is option 'b': 0.2%
Explanation:
The yield strength of a material is defined as the stress in the material when the material begin's to undergo plastic deformation which is also known as yielding.
For materials with well defined yield point such as steel of grade Fe-250 the yield strength can be properly identified from the stress strain curve. But for high strength steel such as Fe-415, Fe-500 the yield point is not properly identified hence the yield strength is taken at 0.2% of proof strain.
The higher the degree of polymerization, the longer the chains will be, which results in a lower chain molecular weight. a)-True b)- false?
Answer:
b)False
Explanation:
When one or more than one monomer join to other monomer then they form a chain and this joining of monomers is called degree of polymerization .
Degree of polymerization :
[tex]Degree\ of\ polymerization=\dfrac{Mass\ of\ polymer}{Mass\ of\ monomer}[/tex]
So from above we can say that when chain will become longer then the weight of polymer will increase.
A particle moves with a constant speed of 6 m/s along a circular path of a radius of 4 m. What is the magnitude of its acceleration. Do not include units in your answer, assumed unit are m/s2.
Answer:
Acceleration in circular path will be 9
Explanation:
We have given speed of the particle in circular path = 6 m/sec
Radius of the circular path = 4 m
We have to find the centripetal acceleration [tex]a_c[/tex]
We know that centripetal acceleration is given by [tex]a_c=\frac{v^2}{r}=\frac{6^2}{4}=9[/tex]
As in question it is given that don't include the unit
So acceleration will be 9
A rectangular sheet of 120 mm x 160 mm can be used to develop the lateral surface of (a) A cylinder of radius 80/π (b) A square prism of side 40 mm (c) A hexagonal prism of side 20 mm (d) All of the above
Answer:
option D is correct
Explanation:
1) for a cylinder with radius is[tex]\frac{80}{\pi}[/tex]
lateral surface area of cylinder is [tex]2\pi rh[/tex]
[tex]= 2\pi \frac{80}{\pi}*120[/tex]
= 160* 120
2) fro square prism with side 40 mm
lateral surface area = 4ah
= 4*40 * 120
= 160*120
3)for hexagonal with side 20 mm
lateral surface area = 6ah
= 6*20*160
=120* 160
therefore option D is correcrt
At the beginning of the compression process of an air-standard Diesel cycle, p1 = 95 kPa and T1 = 300 K. The maximum temperature is 2100 K and the mass of air is 12 g. For a compression ratio of 18, determine the net work developed in kJ (enter a number only)
Answer:
6.8 kJ
Explanation:
p1 = 95 kPa
T1 = 300 K
T3 = 2100 K
m = 12 g
Ideal gas equation:
p * v = R * T
v = R * T / p
R for air is 0.287 kJ/(kg K)
v1 = 0.287 * 300 / 95 = 0.9 m^3/kg
v2 = v1 / cr
v2 = 0.9 / 18 = 0.05 m^3/kg
Assuming an adiabatic compression
p*v^k = constant
k is 1.4 for air
p1 * v1 ^ k = p2 * v2 ^ k
p2 = p1 * (v1 / v2) ^k
p2 = p1 * cr^k
p2 = 95 * 18^1.4 = 5.43 MPa
p1*v1/T1 = p2*v2/T2
T2 = p2*v2*T1/(p1*v1)
T2 = 5430 * 0.05 * 300 / (95 * 0.9) = 952 K
The first principle of thermodynamics
Q = W + ΔU
Since this is an adiabatic process Q = 0
W = -ΔU
W1-2 = -m * Cv * (T2 - T1)
The Cv of air is 0.72 kJ/kg
W1-2 = -0.012 * 0.72 * (952 - 300) = -5.63 kJ
Next the combustion happens and temperature increases suddenly.
v3 = v2 = 0.05 m^3/kg
T2 * p2^((1-k)/k) = T3 * p3^((1-k)/k)
p3 = p2 * (T2/T3)^(k/(1-k)
p3 = 5430 * (952/2100)^(1.4/(1-1.4) = 86.5 MPa
The work is zero because the piston doesn't move.
Next it expands adiabatically:
v4 = v1 = 0.9 m^3/kg
T * v^(k-1) = constant (adiabatic process)
T3 * v3^(k-1) = T4 * v4^(k-1)
T4 = T3 * (v3 / v4)^(k-1)
T4 = 2100 * (0.05 / 0.9)^(1.4-1) = 661 K
p3*v3/T3 = p4*v4/T4
p4 = p3*v3*T4/(v4*T3)
p4 = 86500*0.05*661/(0.9*2100) = 1512 kPa
L3-4 = -m * Cv * (T4 - T3)
L3-4 = -0.012 * 0.72 * (661 - 2100) = 12.43 kJ
Net work:
L1-2 + L3-4 = -5.63 + 12.43 = 6.8 kJ
A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the tangent of the surface of the earth. Assuming the radius of the earth to be 6378 km and that is 5.976 * 10^6 kg, determine the eccentricity of the orbit.
Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore[tex] \theta angle = 0[/tex]
and radial component of given velocity is zero
we have[tex] h = r_o v_r_o = 6378+600 =6.97*10^6 m[/tex]
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
[tex]\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)[/tex]
[tex]GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s[/tex]
so
[tex]\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)[/tex]
solvingt for [tex] \epsilon)[/tex]
[tex]\epsilon = 0.22)[/tex]
therefore eccentrcity of orbit is 0.22
A 29-mm-diameter copper rod is 1.1 m long with a yield strength of 73 MPa. Determine the axial force necessary to cause the diameter of the rod to reduce by 0.01 percent, assuming elastic deformation. Check that the elastic deformation assumption is valid by comparing the axial stress to the yield strength. The axial force necessary to cause the diameter of the rod to reduce by 0.01 percent is ____kN.
Answer:
32.96 MPa
Explanation:
The Poisson ratio of copper is:
μ = 0.355
The Young's modulus of copper is:
E = 117 GPa
The equation for reduction of diameter of a rod is:
D = D0 * (1 - μ*σ/E)
Rearranging:
D = D0 - D0*μ*σ/E
D0*μ*σ/E = D0 - D
D0*μ*σ = E*(D0 - D)
σ = E*(D0 - D) / (D0*μ)
If the diameter is reduced by 0.01 percent
D = 0.9999*D0
σ = E*(D0 - 0.9999*D0) / (D0*μ)
σ = E*(0.0001*D0) / (D0*μ)
σ = 0.0001*E / μ
σ = 0.0001*117*10^9 / 0.355 = 32.96 MPa
This value is below the yield strength, therefore it is valid.
Explain the following boundary layer concepts (i) Boundary layer thickness (ii) Boundary layer transition
Answer with Explanation:
i) Boundary layer thickness: It is the thickness of the boundary layer formed around an object that is placed in the path of a flowing viscous fluid.The boundary layer thickness is the thickness up to which the effect of the object on the flow can be felt. When a viscous flowing fluid encounters an object in it's path of flow, the flowing fluid forms a thin layer of fluid over the object and this layer of fluid is known as boundary layer. This is a phenomenon only observed in the viscous fluids. As shown in the below figure a uniform flow of a viscous fluid encounters a plate, as we can see the thickness of the boundary layer goes on increasing as we move away from the leading edge of the plate the thickness of the boundary layer at any position is termed as boundary layer thickness.
ii) Boundary layer transition: It is the transition of the flow from a laminar nature to fully developed turbulent flow as it moves over an object. It occurs due to change in the Reynolds number of the flow as the effect of boundary layer increases as we move away from the leading edge of the object.
What are factor of safety for brittle and ductile material
Explanation:
Step1
Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.
Step2
It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.
Step3
Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.
The expression for factor of safety for ductile material is given as follows:
[tex]FOS=\frac{\sigma_{yp}}{\sigma_{a}}[/tex]
Here,[tex]\sigma_{f}[/tex] is yield stress and [tex]\sigma_{a}[/tex] is allowable stress.
The expression for factor of safety for brittle material is given as follows:
[tex]FOS=\frac{\sigma_{ut}}{\sigma_{a}}[/tex]
Here,[tex]\sigma_{ut}[/tex] is ultimate stress and [tex]\sigma_{a}[/tex] is allowable stress.
The position of a particle along a straight-line path is defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds. When t = 8 s, determine the particle’s (a) instantaneous velocity and instantaneous acceleration, (b) average velocity and average speed
To determine the instantaneous velocity and acceleration of a particle described by the position function s(t), one needs to calculate the first and second derivatives of the position function and evaluate them at the given time, t = 8 s. The average velocity and speed require the change in position over a specific time interval, which is not provided in the question.
Explanation:The position of a particle along a straight-line path is given by the equation s = (t3 - 6t2 - 15t + 7) feet, where t is time in seconds. To find the particle's instantaneous velocity and instantaneous acceleration at t = 8 s, we need to take the first and second derivatives of the position function with respect to time, respectively.
The first derivative of s with respect to t gives us the velocity v(t), and the second derivative gives us the acceleration a(t). At t = 8 s, the velocities and accelerations can be calculated by plugging in the value of t into those derivatives.
To determine the average velocity and average speed, we take the change in position over the change in time interval for the specific time range provided.
Note: The question lacks sufficient specific information to calculate these values as the time interval for the average velocity and average speed is not provided. However, the general process of calculation has been explained.
Acceleration, instantaneous velocity, and particle position at different times are essential concepts in physics and particle motion analysis.
Acceleration is the rate of change of velocity with respect to time. In the given scenarios, acceleration is provided in terms of a function of time or as a constant value.
Instantaneous velocity is the velocity of the particle at a specific moment. It can be calculated by taking the derivative of the position function with respect to time.
Position of the particle at different times can be found by substituting the respective time values into the position function.